Derivative of ln(cos x): Answer, Proof, Mistakes

The proof: chain rule on ln(cos x)

The outer function is $\ln u$ and the inner is $u = \cos x$. Differentiate the outer to $\frac{1}{u}$, then multiply by the inner derivative $u' = -\sin x$.

The quotient $\frac{\sin x}{\cos x}$ is the definition of $\tan x$, and the minus sign comes straight from the derivative of the inner $\cos x$. This same pattern gives the standard result $\int \tan x\,dx = -\ln|\cos x| + C$ when read in reverse.

Where the domain restricts the formula

The logarithm only accepts positive inputs, so $\ln(\cos x)$ is defined only where $\cos x > 0$, on intervals like $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. The sample points $x = -0.8, 0, 0.5, 1$ all sit inside that interval, where $\cos x$ is positive.

On that interval $-\tan x$ is negative for $x > 0$ and positive for $x < 0$, so $\ln(\cos x)$ rises toward its peak at $x = 0$ and falls after it. That peak matches the maximum of $\cos x$ at $x = 0$.

Common mistakes

• Answering $\tan x$ with no minus sign. The inner derivative of $\cos x$ is $-\sin x$, and that minus carries all the way to the final $-\tan x$.
• Answering $\frac{1}{\cos x}$ and forgetting the chain rule. Differentiating $\ln u$ gives $\frac{1}{u}$, but you still multiply by $u' = -\sin x$.
• Writing $-\frac{\sin x}{\cos x}$ but not recognizing it as $-\tan x$, then mis-simplifying it to $-\sin x \cdot \cos x$.
• Confusing $\ln(\cos x)$ with $\cos(\ln x)$. The order of composition is different, and so is the derivative.